Question

What happens when ethyl alcohol is heated with iodine and $NaOH$?

Answer 1

Alcohols are the chemical compounds and also the most reduced form. Thus, these compounds easily undergo oxidation to form carbonyl compounds and carboxylic acids by the addition of oxidising agents. Iodine with sodium hydroxide forms an oxidising agent sodium iodate, it oxidises ethyl alcohol to acetaldehyde.

Complete answer:
Ethyl alcohol is an alcohol with the molecular formula ${C_2}{H_5}OH$, as it contains a hydroxyl group. It is an alcohol or hydroxyl compound. Alcohols can easily undergo oxidation to carbonyl compounds i.e.., aldehydes or ketones.
Iodine is an oxidising agent and sodium hydroxide is a strong base with the molecular formula of $NaOH$, it is also known as alkali. When sodium hydroxide is treated with iodine, it forms sodium (I) iodate. Sodium (I) iodate is an oxidising agent and oxidises ethyl alcohol to acetaldehyde.
Further, acetaldehyde reacts with sodium (I) iodate and forms sodium formate and iodoform.
The chemical reaction for the formation of ethyl alcohol to iodoform will be as follows:
${C_2}{H_5}OH\xrightarrow{{{I_2}/NaOH}}C{H_3}CHO\xrightarrow{{{I_2}/NaOH}}HCOONa + CH{I_3}$
The formation of an iodoform can be identified from the formation of yellow precipitate which is an iodoform. The chemical formula of iodoform is $CH{I_3}$.

Note:
The formation of iodoform not only indicates the completion of reaction, n=but also indicates the presence of methyl aldehydes or methyl ketones. As the compounds containing methyl groups can only react with iodine to form the precipitate of iodoform. Thus, an iodoform test can also be used to identify the presence of methyl carbonyl compounds.